proof of Hadamard’s inequality

Let’s first prove the second inequality. If $A$ is singular, the thesis is trivially verified, since for a Hermitian positive semidefinite matrix the right-hand side is always nonnegative, all the diagonal entries being nonnegative. Let’s thus assume $det(A)\ne 0$, which means, $A$ being Hermitian positive semidefinite, $det(A)>0$. Then no diagonal entry of $A$ can be $0$ (otherwise, since for a Hermitian positive semidefinite matrix, $0\le {\lambda }_{min}\le a_{ii}\le {\lambda }_{max}$, ${\lambda }_{min}$ and ${\lambda }_{max}$ being respectively the minimal and the maximal eigenvalue, this would imply ${\lambda }_{min}=0$, that is $A$ is singular); for this reason we can define $D=diag(d_{11},d_{22},\mathrm{\dots },d_{nn})$, with $d_{ii}=a_{ii}^{-\frac{1}{2}}\in ℝ$, since all $a_{ii}\in {ℝ}^{+}$. Let’s furthermore define $B=DAD$. It’s easy to check that $B$ too is Hermitian positive semidefinite, so its eigenvalues ${\lambda }_B$ are all non-negative (actually, since $A=A^H$ and since $D$ is real and diagonal, $B^H={(DAD)}^H=D^H{(DA)}^H=D^H{A}^H{D}^H=DAD=B$; on the other hand, for any $𝐱\ne \mathrm{𝟎}$, ${𝐱}^{H}B𝐱={𝐱}^{H}DAD𝐱=({𝐱}^{H}D)A(D𝐱)={(D^H𝐱)}^{H}A(D𝐱)={(D𝐱)}^{H}A(D𝐱)={𝐲}^{H}A𝐲\ge 0$). Moreover, we have obviously $b_{ii}=d_{ii}{a}_{ii}{d}_{ii}=1$ so that $tr(B)=n$ and, recalling the geometric-arithmetic mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality), which holds in this case because the eigenvalues of $B$ are all non-negative,

$det(B)={\prod }_{i=1}^n{\lambda }_B^{(i)}\le {\left(\frac{1}{n}{\sum }_{i=1}^n{\lambda }_B^{(i)}\right)}^n={(\frac{1}{n}tr(B))}^n=1$,

and since $det(B)=det{(D)}^{2}det(A)={\left({\prod }_{i=1}^n{a}_{ii}\right)}^{-1}det(A)$, we have the thesis. Since $det(A)={\prod }_{i=1}^n{a}_{ii}$ if and only if $det(B)=1$ and since in the geometric-arithmetic inequality equality holds if and only if all terms are equal, we must have ${\lambda }_B^{(i)}={\lambda }_B$, so that ${\prod }_{i=1}^n{\lambda }_B^{(i)}={\lambda }_B^n=1$, whence ${\lambda }_B=1$ (${\lambda }_B$ having to be non-negative), and since $B$ is Hermitian and hence is diagonalizable, we obtain $B=I$, and so $A=D^{-1}B{D}^{-1}=D^{-2}=diag(a_{11},a_{22},\mathrm{\dots },a_{nn})$. So we can conclude that equality holds if and only if $A$ is diagonal.

Let’s now derive the more general first inequality. Let $A$ be a complex-valued $n\times n$ matrix. If $A$ is singular, the thesis is trivially verified. Let’s thus assume $det(A)\ne 0$; then $B=A{A}^H$ is a Hermitian positive semidefinite matrix (actually, $B^H={(A{A}^H)}^H={(A^H)}^H{A}^H=A{A}^H=B$ and, for any $𝐱\ne \mathrm{𝟎}$,${𝐱}^{H}B𝐱={𝐱}^{H}A{A}^H𝐱=({𝐱}^{H}A)(A^H𝐱)={(A^H𝐱)}^H(A^H𝐱)={𝐲}^H𝐲={\parallel 𝐲\parallel }_2^2\ge 0$). Therefore, the second inequality can be applied to $B$, yielding:

${\left|det(A)\right|}^2=det(A){det}^{\ast }(A)=det(A)det(A^H)=det(A{A}^H)=$
$=det(B)\le {\prod }_{i=1}^n{b}_{ii}={\prod }_{i=1}^n{\sum }_{j=1}^n{a}_{ij}{(a^H)}_{ji}={\prod }_{i=1}^n{\sum }_{j=1}^n{a}_{ij}{a}_{ij}^{\ast }={\prod }_{i=1}^n{\sum }_{j=1}^n{\left|a_{ij}\right|}^2$.

As we proved above, for $det(B)$ to be equal to ${\prod }_{i=1}^n{b}_{ii}$, $B$ must be diagonal, which means that ${\sum }_{k=1}^n{a}_{ik}{a}_{jk}^{\ast }={|a_{ii}|}^2{\delta }_{ij}$. So we can conclude that equality holds if and only if the rows of $A$ are orthogonal. $\mathrm{\square }$

Title	proof of Hadamard’s inequality
Canonical name	ProofOfHadamardsInequality
Date of creation	2013-03-22 15:37:02
Last modified on	2013-03-22 15:37:02
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	17
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 15A45